solution of neet aipmt

12 Pages·2017·0.25 MB·English

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SOLUTION OF NEET AIPMT - 2017
(HELD ON 7th MAY SUNDAY 2017)

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Set - A (All Code - A/P/W, B/Q/X, C/R/Y, D/S/Z)
BIOLOGY
91. Double fertilization is exhibited by
94. What is the criterion for DNA fragments
(1) Gymnosperms movement on agarose gel during gel
electrophoresis?
(2) Algae
(1) The larger the fragment size, the farther it
(3) Fungi
moves
(4) Angiosperms
(2) The smaller the fragment size, the farther it
Answer (4)
moves
SOLUTION :-
(3) Positively charged fragments move to farther
Double fertilization is a characteristic feature
end
exhibited by angiosperms. It involves syngamy
(4) Negatively charged fragments do not move
and triple fusion.
Answer (2)
92. Which of the following are found in extreme
saline conditions? SOLUTION :-
(1) Archaebacteria During gel electrophoresis, DNA fragments
separate (resolve) according to their size through
(2) Eubacteria
sieving effect provided by agarose gel.
(3) Cyanobacteria
95. Attractants and rewards are required for
(4) Mycobacteria
(1) Anemophily (2) Entomophily
Answer (1)
(3) Hydrophily (4) Cleistogamy
SOLUTION :-
Archaebacteria are able to survive in harsh Answer (2)
conditions because of branched lipid chain in SOLUTION :-
cell membrane which reduces fluidity of cell
Insect pollinated plants provide rewards as edible
membrane. Halophiles (Archaebacteria) are
pollen grain and nectar as usual rewards. While
exclusively found in saline habitats.
some plants also provide safe place for
93. Select the mismatch : deposition of eggs.
(1) Frankia – Alnus 96. Which of the following is made up of dead cells?
(2) Rhodospirillum – Mycorrhiza (1) Xylem parenchyma
(3) Anabaena – Nitrogen fixer (2) Collenchyma
(4) Rhizobium – Alfalfa (3) Phellem
Answer (2) (4) Phloem
SOLUTION :- Answer (3)
Rhodospirillum is anaerobic, free living nitrogen SOLUTION :-
fixer. Mycorrhiza is a symbiotic relationship
Cork cambium undergoes periclinal division and
between fungi and roots of higher plants.
cuts off thick walled suberised dead cork cells
(Phellem) towards outside and it cuts off thin
walled living cells i.e., phelloderm on inner side.
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97. Which cells of ‘Crypts of Lieberkuhn’ secrete DNA as genetic material.
antibacterial lysozyme?
101. Which among the following are the smallest
(1) Argentaffin cells (2) Paneth cells living cells, known without a definite cell wall,
pathogenic to plants as well as animals and can
(3) Zymogen cells (4) Kupffer cells
survive without oxygen?
Answer (2)
(1) Bacillus (2) Pseudomonas
SOLUTION :-
(3) Mycoplasma (4) Nostoc
-Kupffer-cells are phagocytic cells of liver.
– Zymogen cells are enzyme producing cells. Answer (3)
– Paneth cell secretes lysozyme which acts as SOLUTION :-
anti-bacterial agent. Mycoplasmas are smallest, wall-less
prokaryotes, pleomorphic in nature. These are
– Argentaffin cells are hormone producing cells.
pathogenic on both plants and animals.
98. Adult human RBCs are enucleate. Which of the
102. Which of the following options gives the correct
following statement(s) is/are most appropriate
sequence of events during mitosis?
explanation for this feature?
(1) Condensation nuclear membrane
(a) They do not need to reproduce
disassembly crossing over segregation
(b) They are somatic cells
telophase
(c) They do not metabolize
(2) Condensation nuclear membrane
(d) All their internal space is available for oxygen
disassembly arrangement at equator 
transport
centromere division segregation 
(1) Only(d) (2) Only (a)
telophase
(3) (a), (c) and (d) (4) (b) and (c)
(3) Condensation crossing over nuclear
Answer (1)
membrane disassembly segregation 
SOLUTION :- telophase
In Human RBCs, nucleus degenerates during (4) Condensation arrangement at equator
maturation which provide more space for oxygen
centromere division segregation 
carrying pigment (Haemoglobin). It lacks most
telophase
of the cell organelles including mitochondria so
respires anaerobically. Answer (2)
99. The hepatic portal vein drains blood to liver from SOLUTION :-
(1) Heart (2) Stomach The correct sequence of events during mitosis
(3) Kidneys (4) Intestine would be as follows
Answer (4) (i) Condensation of DNA so that chromosomes
become visible occurs during early to mid-
SOLUTION :-
prophase.
In hepatic portal system, hepatic portal vein
carries maximum amount of nutrients from (ii) Nuclear membrane disassembly begins at
intestine to liver. late prophase or transition to metaphase.
100. The final proof for DNA as the genetic material (iii) Arrangement of chromosomes at equator
came from the experiments of occurs during metaphase, called congression.
(1) Griffith (iv) Centromere division or splitting occurs during
(2) Hershey and Chase anaphase forming daughter chromosomes.
(3) Avery, Mcleod and McCarty (v) Segregation also occurs during anaphase as
daughter chromosomes separate and move to
(4) Hargobind Khorana
opposite poles.
Answer (2)
(vi) Telophase leads to formation of two daughter
SOLUTION :-
nuclei.
Hershey and Chase gave unequivocal proof
which ended the debate between protein and
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103. Which one of the following statements is correct, SOLUTION :-
with reference to enzymes? Biosphere reserve is protected area with
(1) Apoenzyme = Holoenzyme + Coenzyme multipurpose activities. It has three zones
(a) Core zone – without any human interference
(2) Holoenzyme = Apoenzyme + Coenzyme
(b) Buffer zone – with limited human activity
(3) Coenzyme = Apoenzyme + Holoenzyme
(c) Transition zone – human settlement, grazing
(4) Holoenzyme = Coenzyme + Cofactor
cultivation etc., are allowed.
Answer (2)
107. A dioecious flowering plant prevents both:
SOLUTION :- (1) Autogamy and xenogamy
Holoenzyme is conjugated enzyme in which (2) Autogamy and geitonogamy
protein part is apoenzyme while non-protein is (3) Geitonogamy and xenogamy
cofactor. Coenzymes are also organic (4) Cleistogamy and xenogamy
compounds but their association with Answer (2)
apoenzyme is only transient and serve as SOLUTION :-
cofactors. When unisexual male and female flowers are
104. During DNA replication, Okazaki fragments are present on different plants the condition is called
used to elongate dioecious and it prevents both autogamy and
(1) The leading strand towards replication fork geitonogamy.
108. A temporary endocrine gland in the human body
(2) The lagging strand towards replication fork
is
(3) The leading strand away from replication fork
(1) Pineal gland (2) Corpus cardiacum
(4) The lagging strand away from the replication
(3) Corpus luteum (4) Corpus allatum
fork
Answer (3)
Answer (4)
SOLUTION :-
SOLUTION :- Corpus luteum is the temporary endocrine
Two DNA polymerase molecules work structure formed in the ovary after ovulation. It
simultaneous at the DNA fork, one on the leading is responsible for the release of the hormones
strand and the other on the lagging strand. like progesterone, oestrogen etc.
Each Okazaki fragment is synthesized by DNA
109. Match the following sexually transmitted
polymerase at lagging strand in 3'5' diseases (Column - I) with their causative agent
direction. (Column - II) and select the correct option.
New Okazaki fragments appear as the Column - I Column- II
replication fork opens further.
(a) Gonorrhea (i) HIV
As the first Okazaki fragment appears away from
(b) Syphilis (ii) Neisseria
the replication fork, the direction of elongation
would be away from replication fork 3'5' (c) Genital Warts (iii) Treponema
direction. (d) AIDS (iv) Human Papilloma
105. Which of the following are not polymeric? virus
(1) Nucleic acids (2) Proteins Options :
(3) Polysaccharides (4) Lipids
(a) (b) (c) (d)
Answer (4)
(1) (ii) (iii) (iv) (i)
SOLUTION :-
(2) (iii) (iv) (i) (ii)
– Nucleic acids are polymers of nucleotides
– Proteins are polymers of amino acids (3) (iv) (ii) (iii) (i)
– Polysaccharides are polymers of (4) (iv) (iii) (ii) (i)
monosaccharides
Answer (1)
– Lipids are the esters of fatty acids and alcohol
SOLUTION :-
106. The region of Biosphere Reserve which is legally
protected and where no human activity is allowed Gonorrhoea – Neisseria (Bacteria) Syphilis –
is known as Treponema (Bacteria)
(1) Core zone (2) Buffer zone Genital Warts – Human papilloma virus (Virus)
(3) Transition zone (4) Restoration zone AIDS – HIV (Virus)
Answer (1)
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110. Transplantation of tissues/organs fails often due 115. The process of separation and purification of
to non-acceptance by the patient’s body. Which expressed protein before marketing is called
type of immune-response is responsible for such (1) Upstream processing
rejections? (2) Downstream processing
(1) Autoimmune response (3) Bioprocessing
(2) Cell-mediated immune response (4) Postproduction processing
Answer (2)
(3) Hormonal immune response
SOLUTION :-
(4) Physiological immune response
Biosynthetic stage for synthesis of product in
Answer (2)
recombinant DNA technology is called
SOLUTION :- upstreaming process while after completion of
Non-acceptance or rejection of graft or biosynthetic stage, the product has to be
transplanted tissues/organs is due to cell subjected through a series of processes which
mediated immune response. include separation and purification are
collectively referred to as downstream
111. Spliceosomes are not found in cells of
processing.
(1) Plants (2) Fungi
116. Mycorrhizae are the example of
(3) Animals (4) Bacteria (1) Fungistasis (2) Amensalism
Answer (4) (3) Antibiosis (4) Mutualism
Answer (4)
SOLUTION :-
SOLUTION :-
Spliceosomes are used in removal of introns
Mycorrhizae is a symbiotic association of fungi
during post-transcriptional processing of hnRNA
with roots of higher plants.
in eukaryotes only as split genes are absent as
117. Viroids differ from viruses in having :
prokaryotes.
(1) DNA molecules with protein coat
112. An example of colonial alga is
(2) DNA molecules without protein coat
(1) Chlorella (2) Volvox
(3) RNA molecules with protein coat
(3) Ulothrix (4) Spirogyra (4) RNA molecules without protein coat
Answer (2) Answer (4)
SOLUTION :- SOLUTION :-
Viroids are sub-viral agents as infectious RNA
Volvox is motile colonial fresh water alga with
particles, without protein coat.
definite number of vegetative cells.
118. Root hairs develop from the region of
113. Which of the following represents order of
(1) Maturation
‘Horse’?
(2) Elongation
(1) Equidae (2) Perissodactyla
(3) Root cap
(3) Caballus (4) Ferus
(4) Meristematic activity
Answer (2) Answer (1)
SOLUTION :- SOLUTION :-
Horse belongs to order perissodactyla of class In roots, the root hairs arise from zone of
mammalia. Perissodactyla includes odd-toed maturation. This zone is differentiated zone thus
mammals. bearing root hairs.
119. Coconut fruit is a
114. Which of the following cell organelles is
(1) Drupe (2) Berry
responsible for extracting energy from
carbohydrates to form ATP? (3) Nut (4) Capsule
Answer (1)
(1) Lysosome (2) Ribosome
SOLUTION :-
(3) Chloroplast (4) Mitochondrion
Coconut fruit is a drupe. A drupe develops from
Answer (4)
monocarpellary superior ovary and are one
SOLUTION :-
seeded.
Mitochondria are the site of aerobic oxidation of
carbohydrates to generate ATP.
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120. Plants which produce characterstic
SOLUTION :-
pneumatophores and show vivipary belong to
The association of H1 protein indicates the
(1) Mesophytes (2) Halophytes
complete formation of nucleosome. Therefore
(3) Psammophytes (4) Hydrophytes the DNA is in condensed form.
Answer (2)
125. DNA fragments are
SOLUTION :-
(1) Positively charged
Halophytes growing in saline soils show
(2) Negatively charged
(i) Vivipary which is in-situ seed germination
(3) Neutral
(ii) Pneumatophores for gaseous exchange
(4) Either positively or negatively charged
121. Which one of the following is related to Ex-situ
depending on their size
conservation of threatened animals and plants?
Answer (2)
(1) Wildlife Safari parks
SOLUTION :-
(2) Biodiversity hot spots
DNA fragments are negatively charged because
(3) Amazon rainforest
of phosphate group.
(4) Himalayan region
126. Capacitation occurs in
Answer (1)
(1) Rete testis
SOLUTION :-
(2) Epididymis
Ex-situ conservation is offsite strategy for
conservation of animals and plants in zoological (3) Vas deferens
park and botanical gardens respectively. (4) Female Reproductive tract
122. Select the mismatch : Answer (4)
(1) Pinus - Dioecious SOLUTION :-
(2) Cycas - Dioecuous Capacitation is increase in fertilising capacity of
(3) Salvinia - Heterosporous sperms which occurs in female reproductive
tract.
(4) Equisetum - Homosporous
127. Which ecosystem has the maximum biomass?
Answer (1)
(1) Forest ecosystem
SOLUTION :-
(2) Grassland ecosystem
Pinus is monoecious plant having both male and
female cones on same plant. (3) Pond ecosystem
123. Which of the following facilitates opening of (4) Lake ecosystem
stomatal aperture? Answer (1)
(1) Contraction of outer wall of guard cells SOLUTION :-
(2) Decrease in turgidity of guard cells High productive ecosystem are
(3) Radial orientation of cellulose microfibrils in – Tropical rain forest
the cell wall of guard cells
– Coral reef
(4) Longitudinal orientation of cellulose
– Estuaries
microfibrils in the cell wall of guard cells
– Sugarcane fields
Answer (3)
128. A disease caused by an autosomal primary non-
SOLUTION :-
disjunction is
Cellulose microfibrils are oriented radially rather
(1) Down’s syndrome
than longitudinally which makes easy for the
stoma to open. (2) Klinefelter’s syndrome
124. The association of histone H1 with a nucleosome (3) Turner’s syndrome
indicates: (4) Sickle cell anemia
(1) Transcription is occurring Answer (1)
(2) DNA replication is occurring SOLUTION :-
(3) The DNA is condensed into a Chromatin
Down’s syndrome is caused by non-disjunction
Fibre
of 21st chromosome.
(4) The DNA double helix is exposed
Answer (3)
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129. Life cycle of Ectocarpus and Fucus respectively (1) Tropical Savannah
are
(2) Tropical Rain Forest
(1) Haplontic, Diplontic
(3) Grassland
(2) Diplontic, Haplodiplontic
(4) Temperate Forest
(3) Haplodiplontic, Diplontic
Answer (2)
(4) Haplodiplontic, Haplontic
SOLUTION :-
Answer (3)
The tropical rain forest have five vertical strata
SOLUTION :- on the basis of height of plants. i.e., ground
Ectocarpus has haplodiplontic life cycle and vegetation, shrubs, short canopy trees, tall
Fucus has diplontic life cycle. canopy trees and tall emergent trees.
130. If there are 999 bases in an RNA that codes for 134. The genotypes of a Husband and Wife are IAIB
a protein with 333 amino acids, and the base at and IAi. Among the blood types of their children
position 901 is deleted such that the length of how many different genotypes and phenotypes
the RNA becomes 998 bases, how many codons are possible
will be altered?
(1) 3 genotypes ; 3 phenotypes
(1) 1 (2) 11
(2) 3 genotypes ; 4 phenotypes
(3) 33 (4) 333
(3) 4 genotypes ; 3 phenotypes
Answer (3)
(4) 4 genotypes ; 4 phenotypes
SOLUTION :-
Answer (3)
If deletion occurs at 901st position the remaining SOLUTION :-
98 bases specifying for 33 codons of amino acids
Husband (X) Wife
will be altered.
131. The pivot joint between atlas and axis is a type (IAIB) (IAi)
of
Possibilities in children
(1) Fibrous joint (2) Cartilaginous joint
IAIB, IAi, IBi, IAIA
(3) Synovial joint (4) Saddle joint
Number of genotypes = 4
Answer (3)
Number of phenotypes = 3
SOLUTION :-
Synovial joints are freely movable joint which IAIA and IAi  A
allow considerable movements. Pivot joint is a
type of synovial joint which provide rotational IAIB= AB
movement as in between atlas and axis
IBi  B
vertebrae of vertebral column.
132. A gene whose expression helps to identify 135. Zygotic meiosis is characteristic of
transformed cell is known as (1) Marchantia (2) Fucus
(1) Selectable marker (3) Funaria (4) Chlamydomonas
(2) Vector Answer (4)
(3) Plasmid SOLUTION :-
(4) Structural gene Chlamydomonas has haplontic life cycle hence
Answer (1) showing zygotic meiosis or initial meiosis.
SOLUTION :- 136. Which of the following is correctly matched for
the product produced by them?
In recombinant DNA technology, selectable
markers helps in identifying and eliminating non- (1) Acetobacter aceti : Antibiotics
transformants and selectively permitting the (2) Methanobacterium : Lactic acid
growth of the transformants.
(3) Penicillium notatum : Acetic acid
133. Presence of plants arranged into well defined
(4) Saccharomyces cerevisiae : Ethanol
vertical layers depending on their height can be
Answer (4)
seen best in :
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SOLUTION :- 140. Which of the following RNAs should be most
Saccharomyces cerevisiae is commonly called abundant in animal cell?
Brewer’s yeast. It causes fermentation of (1) r-RNA (2) t-RNA
carbohydrates producing ethanol.
(3) m-RNA (4) mi-RNA
137. Frog’s heart when taken out of the body
Answer (1)
continues to beat for some time Select the best
option from the following statements SOLUTION :-
(a) Frog is a poikilotherm rRNA is most abundant in animal cell. It
constitutes 80% of total RNA of the cell.
(b) Frog does not have any coronary circulation
141. Which among these is the correct combination
(c) Heart is “myogenic” in nature
of aquatic mammals?
(d) Heart is autoexcitable
(1) Seals, Dolphins, Sharks
Options :
(2) Dolphins, Seals, Trygon
(1) Only (c) (2) Only (d)
(3) Whales, Dolphins, Seals
(3) (a) & (b) (4) (c) & (d)
(4) Trygon, Whales, Seals
Answer (4)
Answer (3)
SOLUTION :-
SOLUTION :-
Frog or the vertebrates have myogenic heart
Sharks and Trygon (sting ray) are the members
having self contractile system or are
of chondrichthyes (cartilaginous fish) while
autoexcitable; because of this condition, it will
whale, Dolphin and Seals are aquatic mammals
keep on working outside the body for some time.
belong to class mammalia.
138. Which statement is wrong for Krebs’ cycle?
142. With reference to factors affecting the rate of
(1) There are three points in the cycle where photosynthesis, which of the following
NAD+ is reduced to NADH + H+ statements is not correct?
(2) There is one point in the cycle where FAD+ (1) Light saturation for CO2 fixation occurs at
is reduced to FADH 10% of full sunlight
(3) During conversion of succinyl CoA to succinic (2) Increasing atmospheric CO2 concentration
acid, a molecule of GTP is synthesised upto 0.05% can enhance CO2 fixation rate
(4) The cycle starts with condensation of acetyl (3) C3 plants responds to higher temperatures
group (acetyl CoA) with pyruvic acid to yield citric with enhanced photosynthesis while C4 plants
acid have much lower temperature optimum
Answer (4) (4) Tomato is a greenhouse crop which can be
SOLUTION :- grown in CO2 - enriched atmosphere for higher
yield
Krebs cycle starts with condensation of acetyl
CoA (2C) with oxaloacetic acid (4C) to form citric Answer (3)
acid (6C). SOLUTION :-
139. In case of poriferans the spongocoel is lined with In C3 plants photosynthesis is decreased at
flagellated cells called : higher temperature due to increased
(1) Ostia photorespiration. C4 plants have higher
temperature optimum because of the presence
(2) Oscula
of pyruvate phosphate dikinase enzyme, which
(3) Choanocytes
is sensitive to low temperature.
(4) Mesenchymal cells
143. Asymptote in a logistic growth curve is obtained
Answer (3) when
SOLUTION :- (1) The value of ‘r’ approaches zero
Choanocytes (collar cells) form lining of (2) K = N
spongocoel in poriferans (sponges). Flagella in
(3) K > N
collar cells provide circulation to water in water
(4) K < N
canal system.
Answer (2)
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SOLUTION :- 145. The DNA fragments separated on an agarose
gel can be visualised after staining with
A population growing in a habitat with limited
resources shows logistic growth curve. For (1) Bromophenol blue
logistic growth
(2) Acetocarmine
dN K N   N  (3) Aniline blue
rN rN 1
    (4) Ethidium bromide
dt  K   K 
Answer (4)
dN SOLUTION :-

where Rate of change in population size,
dt Ethidium bromide is used to stain the DNA
fragments and will appear as orange coloured
r  intrinsic rate of natural increase
bands under UV light.
N  Population size,
146. Functional megaspore in an angiosperm
develops into
K N 
  Environmental resistance. (1) Ovule (2) Endosperm
 K 
(3) Embryo sac (4) Embryo
If K = N then Answer (3)
SOLUTION :-
K N 
0
  , and Megaspore is the first cell of female
 K 
gametophytic generation in angiosperm. It
the population reaches asymptote. undergoes three successive generations of free
nuclear mitosis to form 8-nucleated and 7-celled
144. Out of ‘X’ pairs of ribs in humans only ‘Y’ pairs
embryo sac.
are true ribs. Select the option that correctly
represents values of X and Y and provides their 147. Among the following characters, which one was
explanation : not considered by Mendel in his experiments on
pea?
(1) X = 12, Y = 7 True ribs are attached
(1) Stem – Tall or Dwarf
dorsally to vertebral
(2) Trichomes – Glandular or non-glandular
column and ventrally to
the sternum (3) Seed – Green or Yellow
(2) X = 12, Y = 5 True ribs are attached (4) Pod – Inflated or Constricted
dorsally to vertebral Answer (2)
column and sternum SOLUTION :-
on the two ends During his experiments Mendel studied seven
(3) X = 24, Y = 7 True ribs are dorsally characters. Nature of trichomes i.e., glandular
or non-glandular was not considered by Mendel.
attached to vertebral
148. Lungs are made up of air-filled sacs the alveoli.
column but are free on
They do not collapse even after forceful
ventral side expiration, because of :
(4) X = 24, Y = 12 True ribs are dorsally (1) Residual Volume
attached to vertebral (2) Inspiratory Reserve Volume
column but are free on (3) Tidal Volume
ventral side (4) Expiratory Reserve Volume
Answer (1) Answer (1)
SOLUTION :- SOLUTION :-
In human, 12 pairs of ribs are present in which 7 Volume of air present in lungs after forceful
pairs of ribs (1st to 7th pairs) are attached expiration as residual volume which prevents the
dorsally to vertebral column and ventrally to the collapsing of alveoli even after forceful expiration.
sternum.
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149. GnRH, a hypothalamic hormone, needed in derivative of vitamin A, is the light-absorbing part
reproduction, acts on of all visual photopigments.
(1) Anterior pituitary gland and stimulates 153. Which one of the following statements is not valid
secretion of LH and oxytocin for aerosols?
(2) Anterior pituitary gland and stimulates (1) They are harmful to human health
secretion of LH and FSH
(2) They alter rainfall and monsoon patterns
(3) Posterior pituitary gland and stimulates
(3) They cause increased agricultural productivity
secretion of oxytocin and FSH
(4) They have negative impact on agricultural
(4) Posterior pituitary gland and stimulates
land
secretion of LH and relaxin
Answer (3)
Answer (2)
SOLUTION :-
SOLUTION :-
Aerosols can cause various problems to
Hypothalamus secretes GnRH which stimulates
agriculture through its direct or indirect effects
anterior pituitary gland for the secretion of
on plants. However continually increasing air
gonadotropins (FSH and LH).
pollution may represent a persistent and largely
150. In Bougainvillea thorns are the modifications of irreversible threat to agriculture in the future.
(1) Stipules (2) Adventitious root 154. A decrease in blood pressure/volume will not
cause the release of
(3) Stem (4) Leaf
(1) Renin
Answer (3)
(2) Atrial Natriuretic Factor
SOLUTION :-
(3) Aldosterone
Thorns are hard, pointed straight structures for
protection. These are modified stem (4) ADH
151. Which one from those given below is the period Answer (2)
for Mendel’s hybridization experiments?
SOLUTION :-
(1) 1856 - 1863 (2) 1840 - 1850
A decrease in blood pressure / volume stimulates
(3) 1857 - 1869 (4) 1870 - 1877 the release of renin, aldosterone, and ADH while
increase in blood pressure / volume stimulates
Answer (1)
the release of Atrial Natriuretic Factor (ANF)
SOLUTION :-
which cause vasodilation and also inhibits RAAS
Mendel conducted hybridization experiments on (Renin Angiotensin Aldosterone System)
Pea plant for 7 years between 1856 to 1863 and mechanism that decreases the blood volume/
his data was published in 1865 (according to pressure.
NCERT).
155. Homozygous purelines in cattle can be obtained
152. Good vision depends on adequate intake of by
carotene rich food Select the best option from
(1) mating of related individuals of same breed
the following statements
(2) mating of unrelated individuals of same breed
(a) Vitamin A derivatives are formed from
(3) mating of individuals of different breed
carotene
(4) mating of individuals of different species
(b) The photopigments are embedded in the
membrane discs of the inner segment Answer (1)
(c) Retinal is a derivative of vitamin A SOLUTION :-
(d) Retinal is a light absorbing part of all the visual Inbreeding results in increase in the
photopigments homozygosity. Therefore, mating of the related
individuals of same breed will increase
(1) (a) & (b) (2) (a), (c) & (d)
homozygosity.
(3) (a) & (c) (d) (b), (c) & (d)
156. The vascular cambium normally gives rise to
Answer (2)
(1) Phelloderm (2) Primary phloem
SOLUTION :- (3) Secondary xylem (4) Periderm
Carotene is the source of retinal which is involved Answer (3)
in formation of rhodopsin of rod cells. Retinal, a
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SOLUTION :- (1) Stabilizing selection as it stabilizes this
During secondary growth, vascular cambium character in the population
gives rise to secondary xylem and secondary (2) Directional as it pushes the mean of the
phloem. Phelloderm is formed by cork cambium. character in one direction
157. Which of the following statements is correct?
(3) Disruptive as it splits the population into two
(1) The ascending limb of loop of Henle is one yielding higher output and the other lower
impermeable to water output
(2) The descending limb of loop of Henle is
(4) Stabilizing followed by disruptive as it
impermeable to water
stabilizes the population to produce higher
(3) The ascending limb of loop of Henle is
yielding cows
permeable to water
Answer (2)
(4) The descending limb of loop of Henle is
permeable to electrolytes SOLUTION :-
Answer (1) Artificial selection to obtain cow yielding higher
SOLUTION :- milk output will shift the peak to one direction,
hence, will be an example of Directional
Descending limb of loop of Henle is permeable
selection. In stabilizing selection, the organisms
to water but impermeable to electrolytes while
with the mean value of the trait are selected. In
ascending limb is impermeable to water but
disruptive selection, both extremes get selected.
permeable to electrolytes.
158. Fruit and leaf drop at early stages can be 162. Select the correct route for the passage of
prevented by the application of sperms in male frogs :
(1) Cytokinins (2) Ethylene (1) Testes Bidder’s canal Kidney 
(3) Auxins (4) Gibberellic acid Vasa efferentia Urinogenital duct Cloaca
Answer (3)
(2) Testes Vasa efferentia Kidney 
SOLUTION :-
Seminal Vesicle Urinogenital duct 
Auxins prevent premature leaf and fruit fall. NAA
Cloaca
prevents fruit drop in tomato; 2,4-D prevents fruit
drop in Citrus. (3) Testes Vasa efferentia Bidder’s canal
159. A baby boy aged two years is admitted to play Ureter Cloaca
school and passes through a dental check-up. (4) Testes Vasa efferentia Kidney
The dentist observed that the boy had twenty
Bidder’s canal Urinogenital duct 
teeth. Which teeth were absent?
Cloaca
(1) Incisors (2) Canines
Answer (4)
(3) Pre-molars (4) Molars
SOLUTION :-
Answer (3)
SOLUTION :- In male frog the sperms will move from
Total number of teeth in human child = 20. Testes Vasa efferentia Kidney 
Premolars are absent in primary dentition. Bidder’s canal Urinogenital duct Cloaca.
160. An important characteristic that Hemichordates
163. Which of the following options best represents
share with Chordates is
the enzyme composition of pancreatic juice?
(1) Absence of notochord
(1) Amylase, peptidase, trypsinogen, rennin
(2) Ventral tubular nerve cord
(2) Amylase, pepsin, trypsinogen, maltase
(3) Pharynx with gill slits
(4) Pharynx without gill slits (3) Peptidase, amylase, pepsin, rennin
Answer (3) (4) Lipase, amylase, trypsinogen,
procarboxypeptidase
SOLUTION :-
Answer (4)
Pharyngeal gill slits are present in hemichordates
as well as in chordates. Notochord is present in SOLUTION :-
chordates only. Ventral tubular nerve cord is Rennin and Pepsin enzymes are present in the
characteristic feature of non-chordates. gastric juice. Maltase is present in the intestinal
161. Artificial selection to obtain cows yielding higher juice.
milk output represents
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